[erv10s]

Question:

Starting with 1.0 M stock solutions of H3PO4, NaH2PO4, Na2HPO4, and Na3PO4, describe how to make one liter of a 50 mM sodium phosphate buffer pH 5.5. Assume you have asmuch of any stocks you need. You also have all the water that you need if you need to do a dilution. You cannot add any acid or base to "pH" this --just the right amounts of two of the solutions above. Hint: you may dilute any two solutions you choose to use before you combine them to make your buffer.

My Attempt:

50 mM --> .05 M

pH = pKa + log [A-]/[HA]
5.5 = 7.2 + log [Na2HPO4]/[NaH2PO4]
5.5 = 7.2 + log x/.05-x
-1.7 = log x/.05-x
10^(-1.7) = x/.05-x
.02 = x/.05-x
.02(.05-x) = x
.001 -.02x = x
.001 = 1.02x

x = [Na2HPO4] = 9.8x10^(-4)
.05-x = [NaH2PO4] = .049

I am not sure how to proceed on from this point in terms of figuring out how much stock solution to use and/or if something needs to be diluted. Any help is much appreciated!

[Borek]

Well, you have to be an idiot to make pH 5.5 buffer from phosphates (you are almost 2 units from pKa2, so buffer capacity is very low), but that's not your fault

You are close. You know concentrations. Calculate how much reagents will be present in the final solution, than calculate how much stock solutions contain these amounts of regents. Don't round down intermediate results. Fill up with water.

Note: don't use numbers from the attached picture, they are too exact for your needs - most likely they will be not accepted, as they take into account effects that your calculations ignore. It is just for your reference.

[erv10s]

Just to make sure I understood you correctly:  I used the concentrations that I got to find the moles present in the final solution.

9.8x10^(-4) mol Na2HPO4/L   x 1 L = 9.8x10^(-4) mol
.049 mol NaH2PO4/L  x 1 L = .049 mol

Using those numbers, I calculated the amount of stock solution that would contain that amount.

9.8x10^(-4) mol = (x L)(1 mol Na2HPO4/L)
x = 9.8x10^(-4) L ---> .98 mL

.049 mol = (x L)(1 mol NaH2PO4/L)
x= .049 L --> 49 mL

Then add 950.02 mL water to get the final volume of the buffer.

[Borek]

OK, although I would ignore 950.02 as all you need is to fill up to 1000 mL